Jak zliczyć częstotliwość elementów na liście?
Muszę znaleźć częstotliwość elementów na liście
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
Output - >
b = [4,4,2,1,2]
Również chcę usunąć duplikaty z
a = [1,2,3,4,5]
23 answers
Ponieważ lista jest uporządkowana możesz to zrobić:
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
from itertools import groupby
[len(list(group)) for key, group in groupby(a)]
Wyjście:
[4, 4, 2, 1, 2]
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2010-01-29 12:18:39
W Pythonie 2.7 (lub nowszym) możesz użyć collections.Counter
:
import collections
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counter=collections.Counter(a)
print(counter)
# Counter({1: 4, 2: 4, 3: 2, 5: 2, 4: 1})
print(counter.values())
# [4, 4, 2, 1, 2]
print(counter.keys())
# [1, 2, 3, 4, 5]
print(counter.most_common(3))
# [(1, 4), (2, 4), (3, 2)]
Jeśli używasz Pythona 2.6 lub starszego, możesz go pobrać TUTAJ .
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2017-11-15 22:21:49
Python 2.7 + wprowadza rozumienie słownikowe. Zbudowanie słownika z listy pozwoli Ci policzyć, a także pozbyć się duplikatów.
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> d = {x:a.count(x) for x in a}
>>> d
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
>>> a, b = d.keys(), d.values()
>>> a
[1, 2, 3, 4, 5]
>>> b
[4, 4, 2, 1, 2]
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2012-03-16 20:44:01
Aby policzyć ilość wystąpień:
from collections import defaultdict
appearances = defaultdict(int)
for curr in a:
appearances[curr] += 1
Aby usunąć duplikaty:
a = set(a)
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2010-01-29 12:16:03
Zliczanie częstotliwości elementów najlepiej zrobić za pomocą słownika:
b = {}
for item in a:
b[item] = b.get(item, 0) + 1
Aby usunąć duplikaty, użyj zestawu:
a = list(set(a))
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2010-01-29 12:21:30
W Pythonie 2.7+ możesz użyć kolekcji.Licznik do liczenia elementów
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>>
>>> from collections import Counter
>>> c=Counter(a)
>>>
>>> c.values()
[4, 4, 2, 1, 2]
>>>
>>> c.keys()
[1, 2, 3, 4, 5]
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2010-01-29 13:00:53
seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.
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2010-01-29 12:24:18
Możesz to zrobić:
import numpy as np
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
np.unique(a, return_counts=True)
Wyjście:
(array([1, 2, 3, 4, 5]), array([4, 4, 2, 1, 2], dtype=int64))
Pierwsza tablica to wartości, a druga to liczba elementów z tymi wartościami.
Więc jeśli chcesz uzyskać tylko tablicę z liczbami powinieneś użyć tego:
np.unique(a, return_counts=True)[1]
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2017-07-23 18:17:54
from collections import Counter
a=["E","D","C","G","B","A","B","F","D","D","C","A","G","A","C","B","F","C","B"]
counter=Counter(a)
kk=[list(counter.keys()),list(counter.values())]
pd.DataFrame(np.array(kk).T, columns=['Letter','Count'])
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2017-12-27 22:19:27
Po prostu użyłbym scipy.statystyki.itemfreq w następujący sposób:
from scipy.stats import itemfreq
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
freq = itemfreq(a)
a = freq[:,0]
b = freq[:,1]
Możesz sprawdzić dokumentację tutaj: http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.stats.itemfreq.html
W pierwszym pytaniu zrób iterację listy i użyj słownika, aby śledzić istniejące elementy.
W drugim pytaniu wystarczy użyć operatora set.
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2010-01-29 12:10:59
Ta odpowiedź jest bardziej jednoznaczna
a = [1,1,1,1,2,2,2,2,3,3,3,4,4]
d = {}
for item in a:
if item in d:
d[item] = d.get(item)+1
else:
d[item] = 1
for k,v in d.items():
print(str(k)+':'+str(v))
# output
#1:4
#2:4
#3:3
#4:2
#remove dups
d = set(a)
print(d)
#{1, 2, 3, 4}
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2015-06-30 18:39:12
def frequencyDistribution(data):
return {i: data.count(i) for i in data}
print frequencyDistribution([1,2,3,4])
...
{1: 1, 2: 1, 3: 1, 4: 1} # originalNumber: count
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2016-12-06 16:50:55
from collections import OrderedDict
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
def get_count(lists):
dictionary = OrderedDict()
for val in lists:
dictionary.setdefault(val,[]).append(1)
return [sum(val) for val in dictionary.values()]
print(get_count(a))
>>>[4, 4, 2, 1, 2]
Aby usunąć duplikaty i utrzymać porządek:
list(dict.fromkeys(get_count(a)))
>>>[4, 2, 1]
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2018-01-26 09:26:04
Używam licznika do generowania freq. dict z pliku tekstowego słowa w 1 linijce kodu
def _fileIndex(fh):
''' create a dict using Counter of a
flat list of words (re.findall(re.compile(r"[a-zA-Z]+"), lines)) in (lines in file->for lines in fh)
'''
return Counter(
[wrd.lower() for wrdList in
[words for words in
[re.findall(re.compile(r'[a-zA-Z]+'), lines) for lines in fh]]
for wrd in wrdList])
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2018-01-27 09:37:28
#!usr/bin/python
def frq(words):
freq = {}
for w in words:
if w in freq:
freq[w] = freq.get(w)+1
else:
freq[w] =1
return freq
fp = open("poem","r")
list = fp.read()
fp.close()
input = list.split()
print input
d = frq(input)
print "frequency of input\n: "
print d
fp1 = open("output.txt","w+")
for k,v in d.items():
fp1.write(str(k)+':'+str(v)+"\n")
fp1.close()
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2015-07-10 09:31:01
Jeszcze jedno rozwiązanie z innym algorytmem bez użycia zbiorów:
def countFreq(A):
n=len(A)
count=[0]*n # Create a new list initialized with '0'
for i in range(n):
count[A[i]]+= 1 # increase occurrence for value A[i]
return [x for x in count if x] # return non-zero count
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2015-12-20 08:17:38
num=[3,2,3,5,5,3,7,6,4,6,7,2]
print ('\nelements are:\t',num)
count_dict={}
for elements in num:
count_dict[elements]=num.count(elements)
print ('\nfrequency:\t',count_dict)
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2017-09-03 18:43:08
Możesz użyć wbudowanej funkcji dostarczonej w Pythonie
l.count(l[i])
d=[]
for i in range(len(l)):
if l[i] not in d:
d.append(l[i])
print(l.count(l[i])
Powyższy kod automatycznie usuwa duplikaty na liście, a także wyświetla częstotliwość każdego elementu na liście oryginalnej i listę bez duplikatów.
Dwie pieczenie za jeden strzał ! X DWarning: date(): Invalid date.timezone value 'Europe/Kyiv', we selected the timezone 'UTC' for now. in /var/www/agent_stack/data/www/doraprojects.net/template/agent.layouts/content.php on line 54
2017-10-30 14:49:28
To podejście można wypróbować, jeśli nie chcesz korzystać z żadnej biblioteki i zachować to proste i krótkie!
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
marked = []
b = [(a.count(i), marked.append(i))[0] for i in a if i not in marked]
print(b)
O / P
[4, 4, 2, 1, 2]
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2017-12-06 18:44:18
a=[1,2,3,4,5,1,2,3]
b=[0,0,0,0,0,0,0]
for i in range(0,len(a)):
b[a[i]]+=1
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2018-02-21 03:04:25
Jeszcze jednym sposobem jest użycie słownika i listy.licz, poniżej naiwny sposób na to.
dicio = dict()
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
b = list()
c = list()
for i in a:
if i in dicio: continue
else:
dicio[i] = a.count(i)
b.append(a.count(i))
c.append(i)
print (b)
print (c)
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2017-04-19 19:25:36
str1='the cat sat on the hat hat'
list1=str1.split();
list2=str1.split();
count=0;
m=[];
for i in range(len(list1)):
t=list1.pop(0);
print t
for j in range(len(list2)):
if(t==list2[j]):
count=count+1;
print count
m.append(count)
print m
count=0;
#print m
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2017-02-22 02:18:26